I (asterios) added three more questions besides the infinitness question about this form that someone can maybe answer.  
Consider the following question:

>1) For a given natural number $a$, are there finitely or infinitely many natural numbers that are not of the form $anm \pm n\pm m$, where $m$ and $n$ range over positive integers?  (For $a=1$ or $a=2$ you have all the natural numbers.)

Does this problem appear in the literature? 

As one can see at MO Scribe's question http://mathoverflow.net/questions/49751/chens-theorem-with-congruence-conditions/49782#49782  it is the same like asking if there are infinitely many $k$ such that both $ak+1$ and $ak-1$ do not have any non-trivial factors of the form $\pm 1 \mod a$.

I give a proof that for $a=6$ the question is equivalent to the twin prime conjecture so it is known that we don't have any proof. But what about other values of $a$? Is the problem for $a=100$ or more of the same difficulty?
  
>2) Is the density (Szemeredi's or Schnirelman's), of the numbers that are not of the form, zero for any value of $a$?


>3) From Viggo Brun's theorem we have that the sum of the reciprocals of the twin primes converges. Does the sum of the reciprocals for any value of $a$ of the numbers that are not of this form converge?

>4) Are there infinitely many $k$ such that both  $ak+1$ , $ak-1$ do not have any $prime$ factors  of the form  $\pm 1 \mod a$? (the same questions for these $k$ as in 2) and 3) )

 As MO Scribe noticed the conjectural answer is that there should be infinetely many such pairs because we are aspecting to have infinitely many prime pairs of any reasonable congruence condition.

http://math.stackexchange.com/questions/15075/do-we-have-a-proof-of-the-infiniteness 


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There are infinitely many twin primes if and only if there are infinitely many natural numbers that are not of the form $6nm \pm n \pm m$.

<b>Proof:</b> Every number that is not a multiple of $2$ or $3$ is of the form $6N\pm 1$. So the only pairs that are not divisible by $2$ or $3$ are $(6N-1,6N+1)$ for any $N$. Now are there infinitely many such prime pairs (twin primes)?

If the number $6N-1$ is prime it should not be written as a product of some numbers $6n+1,6m-1$ for any $n,m > 0$. So $(6n+1)(6m-1)=6(6nm-n+m)-1$, which means that $N$ should not be of the form $6nm-n+m$ for any $n,m>0$.

Similarly, if $6N+1$ is a prime it should not be a product of some numbers $(6n-1)(6m-1) =6(6nm-n-m)+1$, or $(6n+1)(6m+1) =6(6nm+n+m)+1$. Which means that we have a prime couple of the form $(6N-1,6N+1)$ if and only if $N$ is not of the form $6nm \pm n \pm m$ for any $n,m$.

NOTE: After i have edited this observasion-question I realised that it is well known that for $a=6$ it is equivalent to twin prime conjecture (as it is written on the answer by Luis below too), a notice that S.Golomb seems to have done first, but my question is focused to the other values of $a$.

If someone want to add something http://tea.mathoverflow.net/discussion/921/reedited/#Item_1